bio | website | |
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Feb 3 |
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Certain conditions on cancellative semigroups
[...] the left-cancellative elements of P(S) are singletons. And the same reasoning can be now dualized to the case of right-cancellative elements, so we see that, at least on the level of groups, (P1) and (P2) are equivalent. (Hope to have added something to what you already knew!) |
Feb 3 |
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Certain conditions on cancellative semigroups
Assume $S$ is a grp (finite or infinite, abelian or not), and suppose to a contradiction that $X$ is left-cancellative for some $X\in P(S)$ with $|X|\ge 2$. Denote by $Y$ the subgrp of $S$ generated by $X$, pick $y\in Y$, and set $Z:=Y\setminus \{y\}$. Claim: $X+Y=X+Z$. Proof. It's enough to show that $X+y\subseteq X+Z$. For, fix $x_1\in X$ and, using $|X|\ge 2$, let $x_2\in X\setminus \{x_1\}$. Then, note that the equ. $x_1y=x_2z$ is solvable in $Z$, as (i) $x_1,x_2,y\in X$, (ii) $Y$ is a subgrp of $S$, and (iii) $x_2^{-1}x_1y\ne y$. This implies the claim, which gives in turn that [...] |
Feb 3 |
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Certain conditions on cancellative semigroups
@Michał. Right, I confused at least a couple of things with others. On a positive note, (P1) and (P2) can be proved to be equivalent if $S$ is a finite group (as far as I can say, you're interested also in the finite case, aren't you?), since then, however we choose $Y,Z\in P(S)$ with $|Y|+|Z| > |S|$, it holds $YZ=S$ (this is a folklore result of additive theory; e.g., it appears as Lemma 3.1.2 in the 3rd chapter of Geroldinger and Ruzsa's Combinatorial NT and Additive Group Theory), so that, if $X\in P(S)$ and $|X|\ge 2$, then $XS = XT = SX = TX = S$ for, say, $T:=S\setminus\{1\}$. |
Feb 3 |
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Certain conditions on cancellative semigroups
If $S$ is a left-cancellative finite sgrp (either commutative or not), then $SX=SS=S$ for every $X\in P(S)$, with the result that no element of $P(S)$ is left-cancellative, unless $S$ is empty or a singleton, in which case (P1) and (P2) are trivially equivalent. This means that, as far as the focus is on left-cancellative sgrps, you're left (excuse the unintended pun!) with the case where no element of $S$, other than the identity if an identity exists, has finite order. |
Jan 24 |
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Isomorphic subcategories of directed graphs and presets
Yes, sorry. I had assumed it was standard terminology. Given a cat $\sf C$, it's said that $\sf D$ is a full subcat of $\sf C$ if, well, $\sf D$ is a subcat of $\sf C$ and $\hom_{\sf D}(X,Y) = \hom_{\sf C}(X,Y)$ for all objects $X,Y \in {\rm Ob}(\sf D)$. |
Jan 24 |
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Isomorphic subcategories of directed graphs and presets
Fixed a couple of adjectives |
Jan 24 |
asked | Isomorphic subcategories of directed graphs and presets |
Sep 24 |
awarded | Autobiographer |
Jul 18 |
awarded | Yearling |
Jul 2 |
awarded | Inquisitive |
Jul 2 |
awarded | Curious |
Apr 30 |
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An introduction to sieve method and their application, Cojocaru & Murty
You should definitely replace the '$x$' appearing as an upper bound for the variable $\delta$ in the expression of the integral $I$ with '$t$'. |
Apr 23 |
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If $(\mathbb M, \tau)$ is a topological monoid, is $\tau$ always induced by a [left] subinvariant semimetric?
I'm finally convinced: there're first-countable topologies which are not semimetrizable (see the comments to mathoverflow.net/questions/163559 for details). |
Apr 23 |
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First-countable topological monoids without local absorbing elements whose topology is induced by a semimetric
[...] About the origins of basic ideas in the area of asymmetric topology'' (in C. E. Aull and R. Lowen (eds.), Handbook of the History of General Topology, Vol. 3, Dordrecht: Kluwer (2001), 853-968), reports a letter by the same Fox where even a paracompact Hausdorff counterexample (to the $\gamma$-space conjecture) is provided. |
Apr 23 |
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First-countable topological monoids without local absorbing elements whose topology is induced by a semimetric
[...] (see Distance function and the metrization problem, BAMS 43 (1937), 133-142). However, it is still true that not all first-countable topologies are semimetrizable, and I learned from R. Fox' work that the question is related to the $\gamma$-space problem (every semimetrizable space is a $\gamma$-space, and it took some time before a disproof of the converse). A Hausdorff counterexample is, in fact, given in: R. Fox, Solution of the $\gamma$-space problem, Proc. AMS 85 (1982), 606-608. And H.-P. A. Künzi, in his survey "Nonsymmetric distances and their associated topologies: [...] |
Apr 23 |
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First-countable topological monoids without local absorbing elements whose topology is induced by a semimetric
Errata corrige. My "proof" that $(X,\tau)$ isn't semimetrizable when $X$ is countably infinite and $\tau$ is the cofinite topology on $X$ was flawed, and in fact, the contrary is true! This follows from a (straightforward) generalization of a theorem by W. A. Wilson dating back to the 1930s, which appears, e.g., as Theorem 6.3.50 in J. Goubault-Larrecq, Non-Hausdorff Topology and Domain Theory: Selected Topics in Point-Set Topology (New Math. Monographs 22, Cambridge Univ. Press, 2013), or can be recovered as an instance of a theorem by A. H. Frink on countably-based quasi-uniformities [...] |
Apr 20 |
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First-countable topological monoids without local absorbing elements whose topology is induced by a semimetric
I'm considering monoids (to me, a monoid is a unital semigroup, or, if you prefer, a unital associative magma in the sense of Bourbaki). But I'm not sure to get the point of your question! What do you mean? Btw, my use of "in addition" in the present formulation of Q2 is misleading: every cancellative monoid is resilient. |
Apr 20 |
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Embedding a linearly ordered free monoid into a linearly ordered group
I will try to dig into all of this, but thanks for sharing your insights! A minor note: I wouldn't say that "bi-ordered" is more conventional than "linearly ordered". The latter appears, e.g., in K. Iwasawa's and P. Conrad's papers. The same Conrad coined (?) the term "o-group" in Extensions of Ordered Groups (Proc. AMS, Vol. 6, No. 4 (Aug., 1955), pp. 516-528), which is still in use by a number of authors. Others do simply talk of "ordered groups" (e.g., G. Freiman and coauthors in the additive theory of groups), and the list continues. But yes, some people prefer "bi-ordered". |
Apr 20 |
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First-countable topological monoids without local absorbing elements whose topology is induced by a semimetric
Sorry for the late reply (I was on holiday). In fact, there're 1st-countable topologies which are not semimetrizable (to be honest, I had no doubt about their existence, but it's only two days ago that I found a counterexample): this is the case, e.g., with the cofinite topology on a countably infinite set. So yes, I restated the OP to take into account your comments. |
Apr 20 |
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First-countable topological monoids without local absorbing elements whose topology is induced by a semimetric
Restated the question to take into account Eric and Chris' comments |